\(\text { The given lines are }\)
\(2 x-y=0 \ldots(1) \)
\(4 x+7 y+5=0 \ldots(2)\)
\(\text { A }(1,2) \text { is a point on line }(1) \text { . }\)
\(\text { Let } \mathrm{B} \text { be the point of intersection of lines }(1) \text { and }(2) .\)
\(\text { On solving equations }(1) \text { and }(2), \text { we obtain } x=\frac{-5}{18} \text { and } y=\frac{-5}{9} \text { . }\)
\(\text { Coordinates of point } \mathrm{B} \text { are }\left(\frac{-5}{18}, \frac{-5}{9}\right) \text { . }\)
\(\text { By using distance formula, the distance between points } \mathrm{A} \text { and } \mathrm{B} \text { can be obtained as }\)
\(\mathrm{AB}=\sqrt{\left(1+\frac{5}{18}\right)^{2}+\left(2+\frac{5}{9}\right)^{2}} \text { units } \)
\(=\sqrt{\left(\frac{23}{18}\right)^{2}+\left(\frac{23}{9}\right)^{2}} \text { units } \)
\(=\sqrt{\left(\frac{23}{2 \times 9}\right)^{2}+\left(\frac{23}{9}\right)^{2}} \text { units } \)
\(=\sqrt{\left(\frac{23}{9}\right)^{2}\left(\frac{1}{2}\right)^{2}+\left(\frac{23}{9}\right)^{2}} \text { units } \)
\(=\sqrt{\left(\frac{23}{9}\right)^{2}\left(\frac{1}{4}+1\right)} \text { units } \)
\(=\frac{23}{9} \sqrt{\frac{5}{4}} \text { units } \)
\(=\frac{23}{9} \times \frac{\sqrt{5}}{2} \text { units } \)
\(=\frac{23 \sqrt{5}}{18} \text { units }\)
\(\text { Thus, the required distance is } \frac{23 \sqrt{5}}{18} \text { units }\)

Let \(\frac{x}{a}+\frac{y}{b}=1\) be the equation of line.
It is given that a + b = 1 and ab = - 6.
We know that (a - b)²  = (a + b)² - 4ab
\(\Rightarrow\) (a - b)² = (1)² - 4 x -6
= 1 + 24 = 25
\(\Rightarrow\) a - b = ±5
Solving a + b = 1 and a - b = 5, we have a = 3 and b = - 2
Solving a + b = 1 and a - b = - 5, we have a = -2 and b = 3
Thus the required equations are
\(\frac{x}{3}+\frac{y}{-2}=1\Rightarrow\) -2x + 3y
= -6 \(\Rightarrow\) 2x - 3y = 6
and  \(\frac{x}{-2}+\frac{y}{3}=1\Rightarrow\) 3x - 2y
= -6 \(\Rightarrow\) -3x + 2y = 6.

Here (x₁, y₁) = (980, 14) and (x₂, y₂) = (1220, 16).
Putting these values in
\(y-y_1=(\frac{y_2-y_1}{x_2-x_1})(x-x_1)\) 
we have
\(y-14=[\frac{16-14}{1220-980}](x-980)\)
\(\Rightarrow y-14=\frac{2}{240}(x-980)\)
\(\Rightarrow y-14=\frac{1}{120}(x-980)\)
\(\Rightarrow 120(y-14)=x-980\)
Putting y = 17, we have
120(17 - 14) = x - 980
\(\Rightarrow\) 120 x 3 = x - 980
\(\Rightarrow\)  x = 1340 litres.

Here X₀ = - 4, Y₀ = 3 and m = \(1\over 2\)
Putting these values in Y - Y₀ = m(x - x₀), we have
\(\therefore y-3=\frac{1}{2}[x-(-4)]\Rightarrow \) 2y - 6 = x + 4
\(\Rightarrow\)  x - 2y + 10 = 0.

The y-coordinate of every point on the x-axis is 0.
Therefore, the equation of the x-axis is y = 0.
The x-coordinate of every point on the y-axis is 0.
Therefore, the equation of the y-axis is x = 0.

Let (a,b) be the coordinates of the foot of the perpendicular from the point (−1,3) to the line 3x−4y−16=0
Slope of the line joining (-1,3) and (a, b), is \(m_1=\frac{b-3}{a+1}\)
Slope of the line 3x−4y−16=0 or \(y=\frac{3}{4} x-4, m_2=\frac{3}{4}\)
Since these two lines are perpendicular, \(\mathrm{m}_1 \mathrm{~m}_2=-1\)
\( \therefore\left\{\frac{b-3}{a+1}\right\} \cdot\left\{\frac{3}{4}\right\}=-1 \)
\( \Rightarrow \frac{3 b-9}{4 a+4}=-1 \)
\( \Rightarrow 3 b-9=4 a-4 \)
\( \Rightarrow 4 a+3 b=5 \ldots(1)\)
Point(a,b) lies on line 3x – 4y = 16
\(\therefore 3 \mathrm{a}-4 \mathrm{~b}=16 \text {....(2) }\)
On solving equation (1) and (2) we obtain
\(\mathrm{a}=\frac{68}{25} \text { and } \mathrm{b}=-\frac{49}{25}\)
Now find the point of intersection of lines (i) and (ii).
\(\left( \frac { 68 }{ 25 } ,\frac { -49 }{ 25 } \right) \).

It is known that the equation of a line whose intercepts on the axes are a and b is
\(\frac{x}{a}+\frac{y}{b}=1 \)
\(\text { or } b x+a y=a b \)
\(\text { or } b x+a y-a b=0\)
\(\text { The perpendicular distance }(d) \text { of a line } A x+B y+C=0 \text { from a point }\left(x_{1}, y_{1}\right) \text { is given by } d=\frac{\left|A x_{1}+B y_{1}+C\right|}{\sqrt{A^{2}+B^{2}}} \text { . }\)
\(\text { On comparing equation (1) to the general equation of line } A x+B y+C=0 \text { , we obtain } A=D, B=a \text { , and } C = -ab\)
\(\text { Therefare, if } p \text { is the length of the perpendicular from point }\left(x_{1}, y_{1}\right)=(0,0) \text { to line }(1) \text { , we abtain }\)
\(p=\frac{|A(0)+B(0)-a b|}{\sqrt{b^{2}+a^{2}}} \)
\(\Rightarrow p=\frac{|-a b|}{\sqrt{a^{2}+b^{2}}}\)
\(\text { On squaring both sides, we abtain }\)
\(p^{2}=\frac{(-a b)^{2}}{a^{2}+b^{2}} \)
\(\Rightarrow p^{2}\left(a^{2}+b^{2}\right)=a^{2} b^{2} \)
\(\Rightarrow \frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{1}{p^{2}} \)
\(\Rightarrow \frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\)
\(\text { Hence, we showed that } \frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}} \text { . }\)

Slope of AB = Slope o fCD
Slope of BC = Slope of DA
Also, slope of AB × slope of BC = −1

Equation of line joining the points  A(3, 0), and B(- 2, - 2) is
\(y-0=\cfrac { -2-0 }{ -2-3 } \left( x-3 \right) \)
 \(\Rightarrow \) 2x - 5y - 6 = 0                                                   
On putting the point C(8, 2) in Eq. (i), we get
\(2\times 8\times -5\times 2-6=0\Rightarrow 0\) , which is true.
Thus, point C also lies on the line.
Hence, the points are collinear.

If the points A (h, 0), B (a, b), and C (0, k) lie on a line, then
Slope of AB = Slope of BC
\(\frac{b-0}{a-h}=\frac{k-b}{0-a}\)
\(\Rightarrow \frac{b}{a-h}=\frac{k-b}{-a}\)
\(\Rightarrow-a b=(k-b)(a-h)\)
\(\Rightarrow-a b=k a-k h-a b+b h\)
\(\Rightarrow k a+b h=k h\)
\(\text { On dividing both sides by } k h, \text { we obtain }\)
\(\frac{k a}{k h}+\frac{b h}{k h}=\frac{k h}{k h}\)
\(\Rightarrow \frac{a}{h}+\frac{b}{k}=1\)
\(\text { Hence, } \frac{a}{h}+\frac{b}{k}=1\)

The vertices of the given triangle are A (4, 4), B (3, 5), and C (–1, –1).
It is known that the slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) 
\(\text { is given by } m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}, x_{2} \neq x_{1} \text { . }\)
\(\text { Slope of } \mathrm{AB}\left(m_{1}\right)=\frac{5-4}{3-4}(-1)\)
\(\text { Slope of } \mathrm{BC}\left(m_{2}\right)=\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}\)
\(\text { Slope of } C A\left(m_{3}\right)=\frac{4+1}{4+1}=\frac{5}{5}=1\)
It is observed that m₁m₃ = –1
This shows that line segments AB and CA are perpendicular to each other
i.e., the given triangle is right-angled at A (4, 4).
Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

Given equation of line is
2x - 3y + 6 = 0
\(\therefore \)  Required distance of the point from the line
= The perpendicular distance from point to the line
\(=\frac { \left| 2\times 2-3(-3)+6 \right| }{ \sqrt { { 2 }^{ 2 }+{ \left( -3 \right) }^{ 2 } } } =\frac { \left| 4+9+6 \right| }{ \sqrt { 4+9 } } =\frac { 19 }{ \sqrt { 13 } } \)

Given, the line intersecting the X-axis to the left of origin. It means it cuts the negative X-axis at a distance of 3 units from the origin.
x- intercept of line on X-axis (d) = -3
Hence, required equation of the line is given by
y = - 2(x + 3) [y = m (x - d)]
y = - 2x - 6    
2x + y + 6 = 0

\(\text { Let } A(x, y) \equiv A(-2-1), B\left(x_{2}, y_{2}\right) \equiv B(4,0)\)\(C(x, y) \equiv C(33) \text { and } D\left(x, y_{j}\right) \equiv(-3,2)\)
Now, mid-point of AC \(=\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right)\)
\(=\left(\frac{-2+3}{2}, \frac{-1+3}{2}\right)=\left(\frac{1}{2}, 1\right)\)...(i)
and mid-point of BD \(=\left(\frac{x_{2}+x_{4}}{2}, \frac{y_{2}+y_{4}}{2}\right)\)
\(=\left(\frac{4-3}{2}, \frac{0+2}{2}\right)=\left(\frac{1}{2}, 1\right)\)  ...(ii)
From Eqs. (i) and (ii), we get Mid-point of AC = Mid-point of BC
Thus, mid-points of both diagonals are coincide each other.
Hence, the points A, B, C and D are vertices of a parallelogram.

30⁰

Use the formula, m = tan θ
-1

(i) The given equation is x + 7y = 0.
It can be written as
\(\Rightarrow 7y=-x\Rightarrow y=\frac{-1}{7}x\)
\(\Rightarrow y = \frac{-1}{7}x+0\)
\(\text { This equation is of the form } y=m x+c, \text { where } m=-\frac{1}{7} \text { and } c=0 \text { . }\)
Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are \(-\frac{1}{7}\) and 0 respectively.
(ii) The given equation is 6x + 3y – 5 = 0
It can be written as
\(y=\frac{1}{3}(-6 x+5) \)
\(y=-2 x+\frac{5}{3}\)
\(\text { This equation is of the form } y=m x+c, \text { where } m=-2 \text { and } c=\frac{5}{3} \text { . }\)
Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are-2  and \(\frac{5}{3}\) respectively.
(iii) The given equation is y = 0.
It can be written as
y = 0.x + 0
This equation is of the form y = mx + c, where m = 0 and c = 0.
Therefore, equation (3) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.

(i) (b)
(ii) (c)
(iii) (a)
(iv) (a)
(v) (c)

(i) (c)
(ii) (b)
(iii) (a)
(iv) (b) 
(v) (c)